Synthesis Discrete-Tim e LTI• Given an open-loop generali zed plant, state equationsxk+1ekyk=A B1B2C1D11D12C2D21D22xkdkuk• Goal: determine if there is a linear controller K,ξk+1uk=AKBKCKDKξkyk(5.11)such that the closed-loop system is– well-defined,– stable, and– has kGedk2←2= kGedk∞< 1.74Fact 1 Characterizing Stability and GainLinear system,xk+1ek=A BC Dxkdk=: MxkdkThen, ρ(A) < 1, max|z|≥1¯σD + C(zI − A)−1B< 1 if and only if there exists0 ≺ W = W∗∈ Fn×nwith¯σW1/2ImA BC DW−1/2Im< 175Fact 2 Constant Matrix Optim ization1. Given R ∈ Fl×l, U ∈ Fl×mand V ∈ Fp×l, where m, p ≤ l.2. Suppose U⊥∈ Fl×(l−m)and V⊥∈ F(l−p)×lhave•U U⊥,VV⊥are both invertible• U∗U⊥= 0m×(l−m), V V∗⊥= 0p×(l−p)3. Let Z ⊂ Fl×lbe a given set of positive definite, Hermitian matricesTheninfQ∈Fm×pZ∈Z¯σhZ1/2(R + UQV ) Z−1/2i< 1if and only if there is a Z ∈ Z such thatV⊥(R∗ZR − Z) V∗⊥≺ 0andU∗⊥RZ−1R∗− Z−1U⊥≺ 0.76Fact 3 Schur ComplementsSuppose that X11∈ Fn×n, Y11∈ Fn×n, with X11= X∗11≻ 0, a nd Y11=Y∗11≻ 0. Let r be a non-negati ve integer. Then there exists matrices X12∈Fn×r, X22∈ Fr×rsuch that X22= X∗22, andX11X12X∗12X22≻ 0 ,X11X12X∗12X22−1=Y11?? ?if and only ifX11InInY11 0⇔ X11 Y−111andrankX11InInY11≤ n + r⇔ rankX11− Y−111≤ rOnto problem...77Synthesis Discrete-Tim e LTI (co nt’d)Write closed-loop system as static feedback around an extended system. Letm denote the (as of yet unknown) state-dimension of the controller. Definean extended plantxk+1ξk+1ekξkyk=A 0 B10 B20 0m0 Im0C10 D110 D120 Im0 0 0C20 D210 D22xkξkdkξk+1ukCall this 5× 5 block matrix Mmss. For the co ntroller in (5.11), define a matrixKssasKss:=AKBKCKDK∈ R(m+nu)×(m+ny)In this notation, the closed-loop system is governedxk+1ξk+1ek= FL(Mmss, Kss)xkξkdk(5.12)FL(Mmss, Kss) isA 0 B10 0m0C10 D11+0 B2Im00 D12KssI −0 00 D22Kss−10 Im0C20 D2178Synthesis Discrete-Tim e LTI (co nt’d)Contro ller achieves stability and kGedk∞< 1 if and only if there exists aW ∈ R(n+m)×(n+m)such that W = WT≻ 0 and¯σW1200 IneFL(Mmss, Kss)W−1200 Ind< 1This is the equation to solve by searching over1. integers m ≥ 0,2. Kss∈ R(m+nu)×(m+ny)and3. W ∈ R(n+m)×(n+m)Without loss of generali ty D22= 0, so definingR :=A 0 B10 0m0C10 D11, U :=0 B2Im00 D12, V :=0 Im0C20 D21givesFL(Mmss, Kss) = R + UKssVClearly, matrices U⊥and V⊥have the formU⊥=U⊥,10U⊥,2, V⊥= [V⊥,10 V⊥,2]where t he matrices U⊥,1, U⊥,2, V⊥,1and V⊥,2are det ermined f rom the data inU and V (could use svd in Matlab).79Synthesis Discrete-Tim e LTI (co nt’d)For notational purposes, defineE :=A B1C1D11Use Y1, Y2and Y3to denote the various entries of the positive definite scalingmatrix W ∈ R(n+m)×(n+m), and X1, X2and X3for W−1,W =Y1Y2YT2Y3W−1=X1X2XT2X3Theorem: There is a controller achieving closed-loop stabi lity and kGedk∞<1if and only if: there is an integer m ≥ 0, a matrix Kss∈ R(m+nu)×(m+ny)and and a matrix 0 ≺ W = WT∈ R(n+m)×(n+m)with Z := diag [W, I]and¯σZ12(R + UKssV ) Z−12< 1if and only if: there is an integer m ≥ 0 , and a matrix 0 ≺ W = WT∈R(n+m)×(n+m)with Z := diag [W, I]UT⊥RZ−1RT− Z−1U⊥≺ 0andV⊥RTZR − ZVT⊥≺ 0Plugging in, and simplifying, gi ves80Synthesis Discrete-Tim e LTI (co nt’d)if and only if: there exist matrices X1, Y1∈ Rn×n, an integer m ≥ 0, withmatrices X2, Y2∈ Rn×m, X3, Y3∈ Rm×m, satisfyingX1X2XT2X3≻ 0 ,X1X2XT2X3−1=Y1Y2YT2Y3such thatUT⊥,1UT⊥,2EX100 IET−X100 IU⊥,1U⊥,2≺ 0andV⊥,1V⊥,2ETY100 IE −Y100 IVT⊥,1VT⊥,2≺ 0Now use Schur complement lemma about extension...81Synthesis Discrete-Tim e LTI: AMI Solutionsif and only if: there exi st matrices X1= XT1≻ 0, Y1= YT1≻ 0 ∈ Rn×n,such thatUT⊥,1UT⊥,2EX100 IET−X100 IU⊥,1U⊥,2≺ 0andV⊥,1V⊥,2ETY100 IE −Y100 IVT⊥,1VT⊥,2≺ 0andX1InInY1 0Remarks:• These equation impose convex constraints on the varia bles {X1, Y1}• Determining the feasibility (or non-feasi bility) of the equations is a con-vex feasibil ity problem• In fact, these expressions ar e all affine in the variables {X1, Y1}, and arecalled affine matrix inequalities (AMI), or just linear matrix inequalities(LMI)• The dimension m necessary to fill out X1and Y1is equal to the rank ofX1− Y−11.• This is at most equal t o n. Hence, if a controller can achieve kGedk∞<1, then without loss i n genera lity, there is a co ntroller with stat e dimen-sion less than or equal to that of the